Ch11 Solutions

668

Solutions to Chapter 11 Problems

11-1 Letting X denote the annual sales of the product, the annual worth for the venture can be determined as

follows:

AW(15%) = $200,000(A/P, 15%, 5) $50,000 0.1($25)X + $12.50X

= $109,660 + $10X

From this equation, we find that X = 10,996 units per year. If it is believed that at least 10,966 units can

be sold each year, the venture appears to be economically worthwhile. Even though the firm does not

know with certainty how many units of the new device will be sold annually, the information provided

by the breakeven analysis will assist management in deciding whether or not to undertake the venture.

Engineering Economy, Fourteenth Edition, by Sullivan, Wicks, and Koelling. ISBN 0-13-614297-4. 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

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669

11-2 The unknown is now the mileage driven each year (instead of fuel cost).

EUACH = $30,000(A/P, 3%, 5) + ($3.50/gal)(X mi/yr)/(30 mpg)

EUACG = $28,000(A/P, 3%, 5) + ($3.50/gal)(X mi/yr)/(25 mpg)

Setting EUACH = EUACG, we find the breakeven mileage to be X = 30,300 miles per year.




670

11-3 The annual operating expenses of long-haul tractors equipped with the various deflectors are calculated

as a function of mileage dirven per year, X:

Windshear: [(X mi/yr)(0.92)(0.2 gal/mi)($3.00/gal)] = $0.552X / yr

Blowby: [(X mi/yr)(0.96)(0.2 gal/mi)($3.00/gal)] = $0.576X / yr

Air-vantage: [(X mi/yr)(0.90)(0.2 gal/mi)($3.00/gal)] = $0.540X / yr

EUAC can now be written in terms of X.

EUACW = $1,000(A/P, 10%, 10) + $10 + $0.552X = $172.70 + $0.552X

EUACB = $400(A/P, 10%, 10) + $5 + $0.576X = $70.08 + $0.576X

EUACA = $1,200(A/P, 10%, 5) + $5 + $0.540X = $321.56 + $0.540X

The breakeven values can be computed between each pair of deflectors by equating their EUAC

equations and solving for X.

Windshear and Blowby: X = 4,276 miles per year

Blowby and Air-vantage: X = 6,986 miles per year

Air-vantage and Windshear: X = 12,405 miles per year

The range over which each deflector is preferred is:

X 4,276 Select Blowby

4,276 X 12,405 Select Windshear

12,405 X Select Air-vantage




671

11-4 The break-even deferment period, T, is determined as follows:

Provide now: PW(costs) = $1,400,000 + $850,000(P/F, 10%, T)

No Provision: PW(costs) = $1,250,000 + $1,150,000(P/F, 10%, T)

If the difference between the two alternatives is examined, it can be seen that $150,000 now is being

traded off against $300,000 at a later date. The question is, what later date constitutes the break-even

point? By equating the PW(costs) and solving, we have 0.5 = (P/F, 10%, T) and T = log(2)/log(1.1) =

7.27. Or, from Appendix C, we see that T is approximately seven years. Thus, if the additional space

will be required in less than seven years, it would be more economical to make immediate provision in

the foundation and structural details. If the addition would not likely be needed until after seven years,

greater economy would be achieved by making no such provision in the first structure.




672

11-5 (a) AW1(15%) = $4,500 (A/P, 15%, 8) + $1,600 $400 + $800 (A/F, 15%, 8)

= $4,500(0.2229) + $1,200 + $800 (0.0729)

= $255

AW2(15%) = $6,000(A/P, 15%,10) + $1,850 $500 + $1,200(A/F,15%,10)

= $6,000(0.1993) + $1,350 + $1,200 (0.0493)

= $213

Therefore, the initial decision is to select Alternative 1. To determine the capital investment of

Alternative 2 (I2) so that the initial decision would be reversed, equate the AWs:

AW1(15%) = AW2(15%)

$255 = I2 (A/P, 15%, 10) + $1,350 + $1,200 (A/F, 15%, 10)

$255 = I2 (0.1993) + $1,350 + $1,200 (0.0493)

I2 = $5,791

The capital investment of Alternative 2 would have to be $5,791 or less for the initial decision to

be reversed.

(b) Set AW1(15%) = AW2(15%) and solve for N assuming market values remain constant.

$4,500 (A/P,15%,N) + $1,200 + $800 (A/F,15%,N) = $213

$4,500 (A/P, 15%, N) + $986.64 + $800 (A/F, 15%, N) = 0

By trial and error, N = 7.3 years.




673

11-6 (a) Let X = operating hours per year

EUACA(12%) = $2,410(A/P, 12%, 8) $80(A/F, 12%, 8)

+ (15 hp)(0.746 kW/hp)($0.06/kWh)(X)/(0.6)

= $478.63 + $1.119X

EUACB(12%) = $4,820(A/P, 12%, 8) + (10 hp)(0.746 kW/hp)($0.06/kWh)(X)/(0.75)

= $970.27 + $0.5968X

Setting EUACA = EUACB we find X = 941 hours per year. At annual operating hours greater than

941 (including 2,000), we prefer the more efficient pump, Pump B. This is confirmed by calculating

the EUAC with X = 2,000.

EUACA(12%) = $2,716 EUACB(12%) = $2,164

(b) Let A = breakeven efficiency of pump A at 2,000 operating hours per year. We already know that

EUACB(12%) = $2,164 at 2,000 hours per year.

EUACA(12%) = $2,164 = $2,410(A/P, 12%, 8) $80(A/F, 12%,8)

+ (15 hp)(0.746 kW/hp)($0.06/kWh)(X)/A

Solving, we get A = 79.67%




674

11-7 Assume repeatability. Set AWA(10%) = AWB(10%) and solve for breakeven value of X.

AWA(10%) = $5,000 (A/P,10%,5) + $1,500 + $1,900 (A/F,10%,5) = $492,22

AWB(10%) = X((A/P,10%,7) + $1,400 + $4,00 (A/F,10%,7)

= 0.2054X + $1,821.60

$492.22 = 0.2054X + $1,821.60

X = ($1,821.60 $492.22) / (0.2054) = $6,472.15

At X = 0, AWB(10%) = $1,821.60 > $492.22, so Alt. B is preferred.

At X = $6,500, AWB(10%) = $486.50 < $492.22, so Alt. A is preferred.

Select Alt. B for 0 X $6,472.15

MARR PWA(MARR) PWB(MARR

)

5% 31k 41.3k

10% 16.3k 21

k

15% 5.5k 5.8

k

20% 2.8k 6k

40k

30

20

10

0

1

0

1

0

0

10

20

30

40

10 0 10 20 30 40

40 30 20 10 0 10

20% 10%

5%

15%

B

A

Polygon Chart of PW

vs. MARR




675

11-8

Alternative A B C

Investment Capital 12,000 15,800 8,000

Annual Savings 4,000 5,200 3,000

MV (after 4 years) 3,000 3,500 1,500

MARR AW(A) AW(B) AW( C) 4% 1,401 1,671 1,149 8% 1,043 1,206 918 12% 677 730 680 16% 304 244 437 20% -77 -251 189




676

11-9 Aftertax, A$ Analysis: Let X = annual beforetax revenue requirement.

EOY

Investment

Annual

Revenu

e

Annual

Expenses

BTCF

(A$)

0 $1,166,0001

$1,166,000

1 X

$519,7503

519,750+X

2 X 545,738 545,738+X

3 X 573,024 573,024+X

4 X 601,676 601,676+X

4 441,7412 441,741

1 Capital Investment = (55 trucks)($21,000/truck) = $1,166,000

2 Market Value = MV4 = 0.35($1,166,000)(1.02)

4 = $441,741

3 Annual Expenses in year k = (55 trucks)(20,000 mi/truck)($0.45/mi)(1.05)

k

= $495,000(1.05)k

EOY

BTCF

(A$)

Depr.

TI

T(38%)

ATCF

(A$)

0 $1,166,000 $1,166,000

1 519,750 + X $388,628 908,378 + X 345,1840.38X 174,566+0.62X

2 545,738 + X 518,287 1,064,025 + X 404,3300.38X 141,408+0.62X

3 573,024 + X 172,685 745,709 + X 283,3690.38X 289,655+0.62X

4 601,676 + X 86,401 688,077 + X 261,4690.38X 340,207+0.62X

4 441,741 441,741 167,862 273,879

PW(15%) = 0 = $1,166,000 $174,566(P/F,15%,1) $141,408 (P/F,15%,2)

$289,655(P/F,15%,3) $340,207(P/F,15%,4)

+ $273,879(P/F,15%,4) + 0.62X(P/A,15%,4)

0 = $1,653,096 + 1.77X

Thus, X = $1,653,096/1.77 = $933,953 in annual revenues per year.

Breakeven Point Interpretation: The equivalent uniform annual revenue of $933,953 per year is the

breakeven point between signing the contract (and purchasing the trucks, etc.), and not signing the

contract (and making no change in current operations).




677

11-10 (a) Annual Revenues = (150 rooms)(0.6)$45

room - day

days

year

365 = $1,478,250

AW(10%) = $1,478,250 $125,000 $5,000,000(A/P,10%,15)

+ (0.2)($5,000,000)(A/F,10%,15) $1,875,000 (A/P,10%,5)

= $232,625 > 0

Yes, the project is economically feasible.

(b) Sensitivity with respect to Decision Reversal

Capital Investment: $232,625* $5,000,000(A/P,10%,15)X = 0

X = 0.3538 or 35.38%

* Market value is assumed to remain constant at $1,000,000.

Occupancy Rate: $232,625 + 45(150)(365)(0.6)X = 0

X = 0.1574 or 15.74%

MARR: (find the IRR and calculate % change)

By trial and error, IRR = 14.3%.

Therefore, the MARR must increase by 14.3/10 1 = 43 %

The decision is most sensitive to changes in occupancy rate (requires the smallest percent change

to reverse the decision).

(c) Annual worth is a linear function with respect to capital investment and occupancy rate we can

construct the plot using points from parts (a) and (b). Annual worth is nonlinear with respect to

the MARR, therefore additional data points are necessary.

i % change AW

6% 40% $557,200

8% 20% 378,559

10% 0 232,625

12% 20% 112,679

14% 40% 12,808




678

11-10 (c) continued

-$600,000

-$400,000

-$200,000

$0

$200,000

$400,000

$600,000

$800,000

$1,000,000

-40 -30 -20 -10 0 10 20 30 40

% Change

An

nu

al

Wo

rth

Capital Investment

MARR

Occupancy Rate




679

11-11 (a) Let X = hours per day for new system.

EUACNew = $150,000(A/P, 1%, 60) $50,000(A/F, 1%, 60) + ($40/hr)(X)(20 days/mo)

EUACUsed = $75,000(A/P, 1%, 60) $20,000(A/F, 1%, 60)

+ ($40/hr)(8 hr/day)(20 day/mo)

Set EUACNew = EUACUsed and solve for X = 6.38 hours per day. This corresponds to an [(8

6.38)/8] 100% = 20.3% reduction in labor hours.

(b) If the new system is expected to reduce labor hours by only 20%, the used system would be

recommended. This conclusion is confirmed by computing the PW of the incremental investment

($75,000) required the new system, PW = $946. But the margin of victory for the used system is

small, and management may elect to go ahead and purchase the new system because of intangible

factors such as reliability and prestige value of having the latest technology.




680

11-12

Most Likely Estimates

New Used

Capital Investment

$150,000 $75,000

Market Value

$50,000 $20,000

Annual Labor Cost

$5,120 $6,400

MARR (per month) 1.00%

Incremental AW

% Change MARR MV (New) Productivity (New)

-40% $ 205 $ (266) $ (778) -30% $ 149 $ (205) $ (589) -20% $ 93 $ (143) $ (399) -10% $ 36 $ (82) $ (210) 0% $ (21) $ (21) $ (21) 10% $ (79) $ 40 $ 168 20% $ (136) $ 101 $ 357 30% $ (195) $ 163 $ 547 40% $ (254) $ 224 $ 736




681

11-13 The entries in the following table are AW values and were generated using the following equation:

AW(6%) = $30,000,000(A/P, 6%, 40) 300($4,000) + occupancy rate rental fee 300

Occupancy Rate

Rental Fee 75% 80% 85% 90% 95% 100%

$6,000 ($1,843,846) ($1,753,846) ($1,663,846) ($1,573,846) ($1,483,846) ($1,393,846)

$7,000 ($1,618,846) ($1,513,846) ($1,408,846) ($1,303,846) ($1,198,846) ($1,093,846)

$8,000 ($1,393,846) ($1,273,846) ($1,153,846) ($1,033,846) ($913,846) ($793,846)

$9,000 ($1,168,846) ($1,033,846) ($898,846) ($763,846) ($628,846) ($493,846)

$10,000 ($943,846) ($793,846) ($643,846) ($493,846) ($343,846) ($193,846)

$11,000 ($718,846) ($553,846) ($388,846) ($223,846) ($58,846) $106,154

$12,000 ($493,846) ($313,846) ($133,846) $46,154 $226,154 $406,154

$13,000 ($268,846) ($73,846) $121,154 $316,154 $511,154 $706,154

$14,000 ($43,846) $166,154 $376,154 $586,154 $796,154 $1,006,154

$15,000 $181,154 $406,154 $631,154 $856,154 $1,081,154 $1,306,154

$16,000 $406,154 $646,154 $886,154 $1,126,154 $1,366,154 $1,606,154

$17,000 $631,154 $886,154 $1,141,154 $1,396,154 $1,651,154 $1,906,154

$18,000 $856,154 $1,126,154 $1,396,154 $1,666,154 $1,936,154 $2,206,154




682

11-14 (a) Analysis at most likely estimates:

PW(15%) = $30,000 + ($20,000 = $5,000) (P/A,15%,5) + $1,000 (P/F,15%,5)

= $20,780.20

Sensitivity to changes in capital investment:

+5%: PW(15%) = $30,000(1.05) _ ($20,000 $5,000)(P/A,15%,5) +

$1,000(P/F,15%,5) = $19,280.20

5%: PW*15%) = $30,000(0.95) + ($20,000 $5,000)(P/A,15%,5) +

$1,000(P/F,15%,5) = $22,280.20

Breakeven percent change:

PW(15%) = 0 = $30,000(1 + x%) + ($20,000 $5,000)(P/A,15%,5) +

$1,000(P/F,15%,5)

x = 0.693 or +69.3% increase in capital investment cost

Sensitivy to changes in annual expenses:

+10%: PW(15%) = $30,000 + [$20,000 $5,000(1.1)](P/A,15%,5) +

$1,000(P/F,15%,5) = $19,104.10

10%: PW(15%) = $30,000 + [$20,000 $5,000)(0.9)](P/A,15%,5) +

$1,000(P/F,15%,5) = $22,456.30


PW(15%) = 0 = $30,000 + [$20,000 $5,000(1 + x%)](P/A,15%,5) +

$1,000(P/F,15%,5)

x = 1.24 or +124% increase in annual expenses

Sensitivity to changes in annual revenue:

+20%: PW(15%) = $30,000 + [$20,000(1.2) $5,000](P/A,15%,5) + $1,000(P/F,15%,5)

= $34,189

20%: PW(15%) = $30,000 + [$20,000(0.8) $5,000](P/A,15%,5) + $1,000(P/F,15%,5)

= $7,371.40


PW(15%) = 0 = $30,000 + [$20,000(1 + x%) = $5,000](P/A,15%,5) + $1,000(P/F,15%,5)

x = 0.31 or 31% (decrease in annual revenues)




683

11-14 (a) continued

Sensitivity to changes in market value:

+20%: PW(15%) = $30,000 + ($20,000 $5,000)(P/A,15%,5) + $1,000(1.2)(P/F,15%,5)

= $20,879.64

20%: PW(15%) = $30,000 + ($20,000 $5,000)(P/A,15%,5) + $1,000(0.8)(P/F,15%,5)

= $20,680.76


PW(15%) = 0 = $30,000 + ($20,000 $5,000)(P/A,15%,5) +

$1,000(1 + x%)(P/F,15%,5)

X = 41.79 or 417.9% (decrease in market value)

Sensitivity to changes in useful life:

At +20%, n = 6

PW(15%) = $30,000 + ($20,000 $5,000)(P/A,15%,6) + $1,000(P/F,15%,6)

= $27,199.80

At 20%, n = 4

PW(15%) = $30,000 + ($20,000 $5,000)(P/A,15%,4) + $1,000(P/F,15%,4)

= $13,396.80


At n = 3 (40%):

PW(15%) = $30,000 + ($20,000 $5,000)(P/A,15%,3) + $1,000(P/F,15%,3)

= $4,905.50

At n =2 (60%):

PW(15%) = $30,000 + ($20,000 $5,000)(P/A,15%,2) + $1,000(P/F,15%,2)

= $4,858.40

Breakeven life 2.5years, a 50% change




684

11-14 (a) continued

Recommendation: Proceed with the project. PW is positive for values of factors (changed individually)

within stated accuracy ranges. However, if all factors are at their worst values, PW(15%) = $1,065

[capital investment at +5%, annual expenses at +10%, annual revenues, market value, and useful life at

20%].

(b) Factors can be ranked base on the breakeven percent change.

Highest need Annual Revenues 31%

Useful Life 50%

Capital Investment +69.3%

Annual Expense +124%

Lowest need Market Value 418%

Percent Devation Changes from Most Likely Estimate

PW

(Thousands

of Dollars)

35

30

25

20

15

10

5

60 50 40 30 20 10 0 10 20 30 40 50 60 70

Capital

Investment

Market

Value

Annual

Revenues

Annual

Expenses

Useful

Life

$20,780




685

11-15 Repair cost = $5,000:

PW(i'%) = 0 = $10,000 + $4,000(P/A,i'%,5) $5,000(P/F,i'%,3)

By trial and error, IRR = 15.5%

Repair cost = $7,000:

PW(i'%) = 0 = $10,000 + $4,000(P/A,i'%,5) $7,000(P/F,i'%,3)

By trial and error, IRR = 9.6%

Repair cost = $3,000

PW(i'%) = 0 = $10,000 + $4,000(P/A,i'%,5) $3,000(P/F,i'%,3)

By trial and error, IRR = 21%

The sensitivity analysis indicates that if the repairs at the end of year 3 cost $5,000 or less, it will be

economical to invest in the machine. However, if the repairs cost $7,000, the IRR of the purchase is less

than the MARR.

A followup analysis would be to determine the maximum repair cost that would still result in the

desired return of 10%. Let R = repair cost at the end of year 3.

PW(10%) = 0 = $10,000 + $4,000(P/A,10%,5) R (P/F,10%,3)

0.7513(R) = $4,163.20

R = $6,872

As long as the repair cost at the end of year 3 does not exceed $6,872 (which represents a 37.44%

increase over the estimated cost of $5,000), the IRR of the purchase will be 10%.




686

11-16 Let x = amount of rebate. Then the purchase price with 2.9% financing is $30,000 and the purchase

price with the 8.9% financing is $30,000 x. We want to find the value of x such that the monthly

payment is the same. Using Excel we can easily find the monthly payment of the 2.9% plan:

PMT(0.029/12, 48, 30,000) = $662.70. Then

30,000 x = PV(0.089/12, 48, 662.70) = $26,681.53

and x = $3,318.47. If you are offered a rebate less than this amount then you should take the 2.9%

financing offer.




687

11-17 1) Assume salvage value for each system = 0

2) The difference in user cost will be projected as a savings for system #2.

Savings = $0.02/vehicle

3) There are approximately 8,760 hours/year

System 1: AW method

CR = $32,000(A/P,10%,10) = $5,206.40

Annual Maintenance = $75

Operation Cost = 78.0

)kWh/08.0)($hr/yr kW)(8,760 28(= $25,157 per year

AC#1 = $5,206.40 + $75 + $25,157 = $30,438.40

System #2:

CR = $45,000(A/P,10%,15) = $5,917.50

Annual Maintenance = $100

Operation Cost = 90.0

)kWh/08.0)($hr/yr kW)(8,760 34(= $26,475

Let N = # vehicles using intersection

Savings per vehicle = ($0.24 $0.22) N = $0.02 N

AC#2 = $5,917.50 + $100 + $26,475 0.2N = $32,492.50 $0.02N

For Breakeven point:

$30,438.40 = $32,492.50 $0.02N and N = 102,705 cars per year

For ADT = Average Daily Traffic

N = 365

705,102= 282 vehicles/day (rounded to next highest integer)




688

11-18 Let X = annual Btu requirement (in 103 Btu)

AWoil (10%) = $80,000(A/P,10%,20) + $4,000 ($2.20/140)(X)

= $5,400 $0.0157X

AWgas (10%) = $60,000(A/P,10%,20) + $6,000 ($0.04/1)(X)

= $1,050 $0.04X

To find breakeven value of X, set AWoil (10%) = AWgas (10%) and solve for X.

$5,400 $0.0157X = $1,050 $0.04X

X = 179,012 or 179,012,000 Btu per year

Now let's examine the sensitivity of the decision to changes in the annual Btu requirement. The

following table and graph indicate that the conversion to natural gas is preferred if the annual Btu

requirement is less than the breakeven point, else the conversion to oil is preferred.

Annual Worth

% change in X Oil Gas

30 $7,430 $6,125

20 7,720 6,850

10 8,010 7,575

0 8,300 8,300

10 8,590 9,025

20 8,880 9,750

30 9,170 10,475

Oil

Gas




689

11-19 (a) AW(optimistic) = $90,000(A/P,10%,12) + $35,000 + $30,000(A/F,10%,12)

= $23,192

AW(most likely) = $100,000(A/P,10%,10) + $30,000 + $20,000(A/F,10%,10)

= $14,984

AW(pessimistic) = $120,000(A/P,10%,6) + $20,000

= $7,552

(b) Net Annual Cash Flow

O M P

Useful O $21,256 $16,256 $6,256

Life M 19,984 14,984 4,984

P 14,632 9,632 368




690

11-20 Build 4lane bridge now:

PW(12%) = $350,000

Build twolane bridge now:

Optimistic: widen bridge to four lanes in 7 years

PW(12%) = $200,000 [$200,000 + (7)($25,000)](P/F,12%,7) = $369,613

Most Likely: widen bridge to four lanes in 5 years

PW(12%) = $200,000 [$200,000 + (5)($25,000)](P/F,12%,5) = $384,405

Pessimistic: widen bridge to four lanes in 4 years

PW(12%) = $200,000 [$200,000 + (4)($25,000)](P/F,12%,4) = $390,650

Recommend building the 4lane bridge now. In this problem, there is no difficulty in interpreting the

results since building the 4lane bridge now is preferred to a delay in widening the bridge for 7 years

(optimistic estimate).

The advantage of pessimistic, most likely, and optimistic estimates is that the uncertainty involved is

made explicit. Therefore, the information should be more useful in decisionmaking. However, when

mixed results are obtained, significant judgement is required in reaching a decision. Mixed results

would occur in this problem, for example, if the PW of a 7year delay in widening the bridge were less

than $350,000.




691

11-21 Let X = miles driven per year.

EUACDart = $13,000(A/P, 10%, 5) + (X/100 mpg)($8/gal)

EUACOther = $10,000(A/P, 10%, 5) + (X/50 mpg)($8/gal)

Setting EUACDart = EUACOther, we can solve for X = 9,892.5 miles per year. If you are planning on

driving 10,000 miles or more per year, the Dart is the most economical vehicle.




692

11-22 (a) Alternative A:

EOY BTCF Depr TI T(40%) ATCF

0 - $108,000,000 --- --- --- - $108,000,000

1 - 3,460,000 $5,400,000 - $8,860,000 $3,544,000 84,000

2 - 3,460,000 10,260,000 - 13,720,000 5,488,000 2,028,000

3 - 3,460,000 9,234,000 - 12,694,000 5,077,600 1,617,600

4 - 3,460,000 8,316,000 - 11,776,000 4,710,400 1,250,400

5 - 3,460,000 7,484,400 - 10,944,400 4,377,760 917,760

6 - 3,460,000 6,728,400 - 10,188,400 4,075,360 615,360

7 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

8 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

9 - 3,460,000 6,382,800 - 9,842,800 3,937,120 477,120

10 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

11 - 3,460,000 6,382,800 - 9,842,800 3,937,120 477,120

12 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

13 - 3,460,000 6,382,800 - 9,842,800 3,937,120 477,120

14 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

15 - 3,460,000 6,382,800 - 9,842,800 3,937,120 477,120

16 - 3,460,000 3,186,000 - 6,646,000 2,658,400 - 801,600

17 - 3,460,000 0 - 3,460,000 1,384,000 - 2,076,000

18 - 3,460,000 0 - 3,460,000 1,384,000 - 2,076,000

19 - 3,460,000 0 - 3,460,000 1,384,000 - 2,076,000

20 - 3,460,000 0 - 3,460,000 1,384,000 - 2,076,000

20 43,200,000 --- 43,200,000 - 17,280,000 25,920,000

PW(10%) = k

0

20

ATCFk (P/F,10%, k) = $99,472,154




693

11-22 (a) continued

Alternative B: Compute ATCFs for current estimate of capital investment.

Using the ATCFs shown in the following table:

PW(10%) = k

0

20

ATCFk (P/F,10%, k) = $79,065,532

ATCFs for Alternative B given original capital investment amount:


0 - $17,000,000 --- --- --- - $17,000,000

1 - 12,400,000 850,000$ - $13,250,000 $5,300,000 - 7,100,000

2 - 12,400,000 1,615,000 - 14,015,000 5,606,000 - 6,794,000

3 - 12,400,000 1,453,500 - 13,853,500 5,541,400 - 6,858,600

4 - 12,400,000 1,309,000 - 13,709,000 5,483,600 - 6,916,400

5 - 15,400,000 1,178,100 - 16,578,100 6,631,240 - 8,768,760

6 - 12,400,000 1,059,100 - 13,459,100 5,383,640 - 7,016,360

7 - 12,400,000 1,003,000 - 13,403,000 5,361,200 - 7,038,800

8 - 12,400,000 1,003,000 - 13,403,000 5,361,200 - 7,038,800

9 - 12,400,000 1,004,700 - 13,404,700 5,361,880 - 7,038,120

10 - 15,400,000 1,003,000 - 16,403,000 6,561,200 - 8,838,800

11 - 12,400,000 1,004,700 - 13,404,700 5,361,880 - 7,038,120

12 - 12,400,000 1,003,000 - 13,403,000 5,361,200 - 7,038,800

13 - 12,400,000 1,004,700 - 13,404,700 5,361,880 - 7,038,120

14 - 12,400,000 1,003,000 - 13,403,000 5,361,200 - 7,038,800

15 - 15,400,000 1,004,700 - 16,404,700 6,561,880 - 8,838,120

16 - 12,400,000 501,500 - 12,901,500 5,160,600 - 7,239,400

17 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

18 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

19 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

20 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

20 0 --- 0 0 0




694

11-22 (a) continued

Alternative B (revised to include extra investment permissible to breakeven)


0 - $42,731,490 --- --- --- - $42,731,490

1 - 12,400,000 $2,136,574 - $14,536,574 $5,814,630 - 6,585,370

2 - 12,400,000 4,059,492 - 16,459,492 6,583,797 - 5,816,203

3 - 12,400,000 3,653,542 - 16,053,542 6,421,417 - 5,978,583

4 - 12,400,000 3,290,325 - 15,690,325 6,276,130 - 6,123,870

5 - 15,400,000 2,961,292 - 18,361,292 7,344,517 - 8,055,483

6 - 12,400,000 2,662,172 - 15,062,172 6,024,869 - 6,375,131

7 - 12,400,000 2,521,158 - 14,921,158 5,968,463 - 6,431,537

8 - 12,400,000 2,521,158 - 14,921,158 5,968,463 - 6,431,537

9 - 12,400,000 2,525,431 - 14,925,431 5,970,172 - 6,429,828

10 - 15,400,000 2,521,158 - 17,921,158 7,168,463 - 8,231,537

11 - 12,400,000 2,525,431 - 14,925,431 5,970,172 - 6,429,828

12 - 12,400,000 2,521,158 - 14,921,158 5,968,463 - 6,431,537

13 - 12,400,000 2,525,431 - 14,925,431 5,970,172 - 6,429,828

14 - 12,400,000 2,521,158 - 14,921,158 5,968,463 - 6,431,537

15 - 15,400,000 2,525,431 - 17,925,431 7,170,172 - 8,229,828

16 - 12,400,000 1,260,579 - 13,660,579 5,464,232 - 6,935,768

17 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

18 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

19 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

20 - 12,400,000 0 - 12,400,000 4,960,000 - 7,440,000

20 0 --- 0 0 0

The above solution for Alternative B is the result of a trial and error procedure for obtaining

identical present worths of ATCFs.

Extra Capital = $42,731,490 $17,000,000 = $25,731,490

This solution takes into account depreciation credits arising from the extra capital that can be

invested in Alternative B to breakeven with Alternative A.




695

11-22 (b) Coterminate both alternatives at the end of year 10.

Alternative A:


0 - $108,000,000 --- --- --- - $108,000,000

1 - 3,460,000 $5,400,000 - $8,860,000 $3,544,000 84,000

2 - 3,460,000 10,260,000 - 13,720,000 5,488,000 2,028,000

3 - 3,460,000 9,234,000 - 12,694,000 5,077,600 1,617,600

4 - 3,460,000 8,316,000 - 11,776,000 4,710,400 1,250,400

5 - 3,460,000 7,484,400 - 10,944,400 4,377,760 917,760

6 - 3,460,000 6,728,400 - 10,188,400 4,075,360 615,360

7 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

8 - 3,460,000 6,372,000 - 9,832,000 3,932,800 472,800

9 - 3,460,000 6,382,800 - 9,842,800 3,937,120 477,120

10 - 3,460,000 3,186,000 - 6,646,000 2,658,400 - 801,600

10 43,200,000 --- 4,935,600 - 1,974,240 41,225,760

*MV10 = $43,200,000; BV10 = $38,264,400

PW(10%) =

k

0

10

ATCFk (P/F,10%, k) = $87,010,230

Alternative B:


0 - $17,000,000 --- --- --- - $17,000,000

1 - 12,400,000 $850,000 - $13,250,000 $5,300,000 - 7,100,000

2 - 12,400,000 1,615,000 - 14,015,000 5,606,000 - 6,794,000

3 - 12,400,000 1,453,500 - 13,853,500 5,541,400 - 6,858,600

4 - 12,400,000 1,309,000 - 13,709,000 5,483,600 - 6,916,400

5 - 15,400,000 1,178,100 - 16,578,100 6,631,240 - 8,768,760

6 - 12,400,000 1,059,100 - 13,459,100 5,383,640 - 7,016,360

7 - 12,400,000 1,003,000 - 13,403,000 5,361,200 - 7,038,800

8 - 12,400,000 1,003,000 - 13,403,000 5,361,200 - 7,038,800

9 - 12,400,000 1,004,700 - 13,404,700 5,361,880 - 7,038,120

10 - 15,400,000 501,500 - 15,901,500 6,360,600 - 9,039,400

10 0 --- - 6,023,100 2,409,240 2,409,240

*MV10 = 0; BV10 = $6,023,100

PW(10%) =

k

0

10

ATCFk (P/F,10%, k) = $60,788,379




696

11-22 (b) continued

If the study period is reduced to 10 years, Alternative B would still be recommended.

Sensitivity of Alternative B to cotermination at EOY 10:

%100

532,065,79$

379,788,60$532,065,79$ 23.1% less expensive.

(c) If our annual operating expenses of Alternative B double, the extra present worth of cost in part (a)

equals:

($2.1 million)(1 0.4) (P/A, 10%, 20) = $10,727,090.

This makes the total present worth of Alternative B equal to:

$79,065,532 $10,727,090 = $89,792,622.

Because $89,792,622 > $99,472,154, the initial decision to adopt Alternative B is not reversed.

EOY BTCF Depr TI T (40%) ATCF

0 - $17,000,000 --- --- --- - $17,000,000

1 - 14,500,000 850,000$ - $15,350,000 $6,140,000 - 8,360,000

2 - 14,500,000 1,615,000 - 16,115,000 6,446,000 - 8,054,000

3 - 14,500,000 1,453,500 - 15,953,500 6,381,400 - 8,118,600

4 - 14,500,000 1,309,000 - 15,809,000 6,323,600 - 8,176,400

5 - 17,500,000 1,178,100 - 18,678,100 7,471,240 - 10,028,760

6 - 14,500,000 1,059,100 - 15,559,100 6,223,640 - 8,276,360

7 - 14,500,000 1,003,000 - 15,503,000 6,201,200 - 8,298,800

8 - 14,500,000 1,003,000 - 15,503,000 6,201,200 - 8,298,800

9 - 14,500,000 1,004,700 - 15,504,700 6,201,880 - 8,298,120

10 - 17,500,000 1,003,000 - 18,503,000 7,401,200 - 10,098,800

11 - 14,500,000 1,004,700 - 15,504,700 6,201,880 - 8,298,120

12 - 14,500,000 1,003,000 - 15,503,000 6,201,200 - 8,298,800

13 - 14,500,000 1,004,700 - 15,504,700 6,201,880 - 8,298,120

14 - 14,500,000 1,003,000 - 15,503,000 6,201,200 - 8,298,800

15 - 17,500,000 1,004,700 - 18,504,700 7,401,880 - 10,098,120

16 - 14,500,000 501,500 - 15,001,500 6,000,600 - 8,499,400

17 - 14,500,000 0 - 14,500,000 5,800,000 - 8,700,000

18 - 14,500,000 0 - 14,500,000 5,800,000 - 8,700,000

19 - 14,500,000 0 - 14,500,000 5,800,000 - 8,700,000

20 - 14,500,000 0 - 14,500,000 5,800,000 - 8,700,000

20 0 --- 0 0 0




697

Solutions to Spreadsheet Exercises

11-23 Left as an individual exercise. See F11-6.xls.




698

11-24 See P11-24.xls

Single Factor Change: Fuel Economy of Gas Engine (mpg)

24 25 26 27

Extra Cost $1,200 12% $287.41 $262.60 $239.69 $218.49

$ / gal $3.00 MARR 13% $279.12 $254.31 $231.41 $210.20

Miles/yr 20,000 14% $270.76 $245.95 $223.05 $201.84

15% $262.32 $237.51 $214.61 $193.40

Fuel Economy (mpg):

Gas engine 25

% Improvement 33%

Diesel engine 33.25

MARR 14%

AW(fuel savings) 595.49$

Net AW 245.95$




699

11-25 See P11-25.xls.

MARR 20% Useful Life 20

Installation

Expense ($/in.) 150

Operating Hours

per Year 8,760

Annual Tax and

Insurance Rate 5%

Cost of Heat Loss

($/Btu) 0.00004

Insulation

Thickness (in.)

Heat Loss

(Btu/hr)

Installation

Expense

Annual Taxes and

Insurance

Cost of Heat

Removal ($/yr.)

Equiv. Uniform

Annual Cost

3 4,400 450 22.5 1541.76 $1,656.67

4 3,400 600 30.0 1191.36 $1,344.57

5 2,800 750 37.5 981.12 $1,172.64

6 2,400 900 45.0 840.96 $1,070.78

7 2,000 1050 52.5 700.80 $968.92

8 1,800 1200 60.0 630.72 $937.15

Change in Cost Equivalent Uniform Annual Cost

of Heat Loss 3 4 5 6 7 8

-50% $885.79 $748.89 $682.08 $650.30 $618.52 $621.79

-40% $1,039.97 $868.03 $780.19 $734.40 $688.60 $684.86

-30% $1,194.14 $987.17 $878.30 $818.49 $758.68 $747.93

-20% $1,348.32 $1,106.30 $976.41 $902.59 $828.76 $811.00

-10% $1,502.49 $1,225.44 $1,074.53 $986.68 $898.84 $874.08

0% $1,656.67 $1,344.57 $1,172.64 $1,070.78 $968.92 $937.15

10% $1,810.85 $1,463.71 $1,270.75 $1,154.88 $1,039.00 $1,000.22

20% $1,965.02 $1,582.85 $1,368.86 $1,238.97 $1,109.08 $1,063.29

30% $2,119.20 $1,701.98 $1,466.97 $1,323.07 $1,179.16 $1,126.36

40% $2,273.37 $1,821.12 $1,565.09 $1,407.16 $1,249.24 $1,189.44

50% $2,427.55 $1,940.25 $1,663.20 $1,491.26 $1,319.32 $1,252.51




700

11-25 continued

MARR 20% Useful Life 20Installation Expense ($/in.) 150$

Operating Hours per Yr. 8,760

Annual Tax and Insurance Rate 5%

Cost of Heat Loss ($/Btu) 0.00002$

Insulation Thickness (in.)

Heat Loss (Btu/hr)

Installation Expense

Annual Taxes and Insurance

Cost of Heat Removal ($/yr)

Equivalent Annual Worth

3 4,400 (450)$ (22.50)$ (770.88)$ (885.79)$ 4 3,400 (600)$ (30.00)$ (595.68)$ (748.89)$ 5 2,800 (750)$ (37.50)$ (490.56)$ (682.08)$ 6 2,400 (900)$ (45.00)$ (420.48)$ (650.30)$ 7 2,000 (1,050)$ (52.50)$ (350.40)$ (618.52)$ 8 1,800 (1,200)$ (60.00)$ (315.36)$ (621.79)$

Equivalent Annual Worth3 4 5 6 7 8

-50% (500.35)$ (451.05)$ (436.80)$ (440.06)$ (443.32)$ (464.11)$ -40% (577.44)$ (510.62)$ (485.85)$ (482.11)$ (478.36)$ (495.64)$ -30% (654.53)$ (570.19)$ (534.91)$ (524.16)$ (513.40)$ (527.18)$ -20% (731.61)$ (629.76)$ (583.97)$ (566.20)$ (548.44)$ (558.72)$ -10% (808.70)$ (689.33)$ (633.02)$ (608.25)$ (583.48)$ (590.25)$ 0% (885.79)$ (748.89)$ (682.08)$ (650.30)$ (618.52)$ (621.79)$ 10% (962.88)$ (808.46)$ (731.13)$ (692.35)$ (653.56)$ (653.32)$ 20% (1,039.97)$ (868.03)$ (780.19)$ (734.40)$ (688.60)$ (684.86)$ 30% (1,117.05)$ (927.60)$ (829.25)$ (776.44)$ (723.64)$ (716.40)$ 40% (1,194.14)$ (987.17)$ (878.30)$ (818.49)$ (758.68)$ (747.93)$ 50% (1,271.23)$ (1,046.73)$ (927.36)$ (860.54)$ (793.72)$ (779.47)$




701

11-26 See P11-26.xls.

O ML P

Capital Investment $90,000 $100,000 $120,000

Useful Life (years) 12 10 6

Market Value $30,000 $20,000 $0

Net annual cash flow $35,000 $30,000 $20,000

MARR 11% 11% 11%

Annual Worth $22,458 $14,216 -$8,365

Net Annual Cash Flow

O ML P

Useful Life $35,000 $30,000 $20,000

O 12 $20,478 $15,478 $5,478

ML 10 $19,216 $14,216 $4,216

P 6 $13,890 $8,890 ($1,110)




702

11-27 See P11-27.xls.

Most Likely Estimates: % Change Inv. MV Ann. Sav.

-40% $4,852,706 ($290,801) ($2,344,870)

Investment $10,000,000 -30% $3,852,706 ($4,924) ($1,545,476)

Market Value $5,000,000 -20% $2,852,706 $280,952 ($746,082)

Annual Savings $2,800,000 -10% $1,852,706 $566,829 $53,312

0% $852,706 $852,706 $852,706

MARR 15% 10% ($147,294) $1,138,582 $1,652,100

20% ($1,147,294) $1,424,459 $2,451,494

Present Worth $852,706 30% ($2,147,294) $1,710,336 $3,250,887

40% ($3,147,294) $1,996,212 $4,050,281

Breakeven Points:

Investment $10,852,706

Market Value $5,000,000

Annual Savings $2,800,000

MARR 15%

Present Worth $0.00




MARR 15%

Present Worth $0




MARR 15%

Present Worth $0




703

Solutions to Case Study Exercises

11-28 The following sensitivity graph was created using the results from the case study:

-$300,000

-$200,000

-$100,000

$0

$100,000

$200,000

$300,000

$400,000

-40 -30 -20 -10 0 10 20 30 40

Percent Change in Factor

AW

(15%

)

Raw Material Costs: Assume, because of competition, that only 50% of any material cost increase

(above $27/yd3) can be recovered through an increase in the selling price ($45/yd

3). The resulting

AW(15%) values with 10, 20, and 30% material cost increases are:

Percent Increase in Material Cost (M)

10 20 30

AW(15%) $35,397 $17,172 $1,053

Material Cost

Useful Life Capacity

Utilization

Selling Price




704

11-29 The three most sensitive factors are: capacity utilization (U), selling price (S), and material costs (M).

Two approaches are used in this solution. The first involves a modified O ML P approach to

develop scenarios. The second uses selected changes in the three factors, in combination, for the

development of scenarios.

Modified O ML P scenarios

Estimated Value

Factor O* ML

* P

Selling Price (S) $45 $45 $40.50 (10%)

Capacity Utilization (U) 90% 75% 50%

Material Costs (M) $27 $27 $35.10 (+30%)

* For S and M the optimistic and most likely estimates are the same. The three factors have 2 2

3 = 12 combinations or scenarios.

AW(15%) = $72(250)(U) SM 27

2

Annual Revenue

$72(250)(U)(M) Material costs

$9,143(1 + U) O&M expenses

$182,521 Other expenses

= $18,000(U) SM 27

2M

$9,143(U) $182,521

* Assumes that 50% of any material cost increase above $27/yd

3 can be recovered through an

increase in the selling price ($45/yd3 or $40.50/yd

3).

Selling Price (S)

O, ML: $45 P: $40.50

Capacity Material Cost (M)

Utilization (U) O, ML: $27 P: $35.10 O, ML: $27 P: $35.10

O: 90% $101* $35 $28 $38

ML: 75% 54 1 7 62

P: 50% 25 62 66 102

* AW(15%) of scenario (in thousands) based on calculation above.

This modified O ML P analysis highlights the importance of U > 75% and material costs (M)

remaining close to $27/yd3. The project results are sensitive to combinations of changes in these

factors.




705

11-29 continued

Other Selected Scenarios

Percent Deviation for Combination:

Factor A B C

Capacity Utilization 10% 15% 25%

Material Cost +10% +10% + 5%

Selling Price 0% + 3% 0%

AW(15%) $2,800 $2,850 $18,940

These additional scenarios further emphasize the sensitivity of the project to the combined effect

of modest changes in the three factors.




706

Solutions to FE Practice Problems

11-30 Existing Bridge:

PWE(12%) = $1,6000,000 $20,000(P/A,12%,20) $70,000[(P/F,12%,5)

+(P/F,12%,10) + (P/F,12%,15)] = $1,824,435

New Bridge:

PWN(12%) = X [$24,000 + (5)($10,000)](P/A,12%,20) +

($0.25)(4000,000)(P/A,12%,20)

= X + $194,204

Set PWE(12%) = PWN(12%) and solve for X.

$1,824,435 = X + $194,204

X = $2,018,639

Select (b)




707

11-31 X = average number of vehicles per day

AW(12%) = 0 = $117,000/mile(A/P,12%,25) mile

)000,117)($03.0(

+ (X vehicles/day)(365 days/yr)($1,200/accident)

milesvehicle

accidents

000,000,1

710250,1

X = 77.91 vehicles/day

Select (a)




708

11-32 AW(12%) = $8,000(A/P,12%,7) + X($0.50 $0.26) $2,000 = 0

X = 15,637

Select (e)




709

11-33 AW(12%) = $16,000(A/P.12%,7) + X($0.50 $0.16) $4,000 = 0

X = 22,076

Select (c)




710

11-34 AWA(12%) $8,000(A/P,12%,7) + 35,000($0.50 $0.26) $2,000 = $4,647

AWB(12%) = $16,000(A/F,12%,7) + 35,000($0.50 $0.16) $4,000 = $4,394

Select (b) Install Machine A




711

11-35 False




712

11-36 False




713

11-37 False




714

11-38 True




715

11-39 False




Ch11 Solutions

Documents